Step by step
The question
Question: A 13 ft ladder leans against a wall.
The top slides down at 3 ft/s.
How fast is the bottom sliding away when the top is 12 ft above the ground?
1 Draw and name
Draw it.
The ladder stays 13 ft long.
The bottom's distance from the wall, x, and the top's height, y, change.
The top slides down: y gets smaller by 3 ft per second.
Getting smaller means a minus rate: dy/dt = −3.
dx/dt means how fast x changes: the change each second.
We want dx/dt at the moment when y = 12.
2 Write the equation
The wall, the ground and the ladder make a right triangle.
Pythagoras: x squared plus y squared is 13 squared.
Careful!
Do not put y = 12 in yet!
y changes with time.
Numbers go in only after the derivative.
3 Take the derivative
Both x and y change with time.
The chain rule: x squared gives 2x times dx/dt.
y squared gives 2y times dy/dt.
13 squared is a number: 0.
4 Put the numbers in
First the missing side at that moment: x.
Pythagoras with y = 12.
13 squared is 169, 12 squared is 144.
Take away: 25.
So x = 5.
A length is positive.
Now put in x = 5, y = 12, and dy/dt = −3.
Multiply out.
5 Solve for the rate
Add 72 on both sides.
To remove times 10, we divide by 10.
So dx/dt = 7.2 ft per second.
The bottom slides away at 7.2 ft per second.
6 Check
Let's check: a tiny moment later, 0.001 seconds.
y is 11.997.
Then x, from Pythagoras, is about 5.00719.
x changed by 0.00719 in 0.001 seconds.
That is about 7.19 ft per second.
Right!
Let's look back.
First: draw and name.
Then: write the equation.
Then: take the derivative.
Then: put the numbers in.
Then: solve for the rate.
Last: check.